15b - ACI Mix Design - ACI Mix Design (Updated).pdf · ACI Mix Design We’ll work through the mix...

ACI Mix Design

Updated Version

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ACI Mix Design

So-called “mix design” methods actually produce afirst guess at the proper mix proportions. That trialmix is then made in the lab and tested for slump,strength and other pertinent properties and the mixproportions are adjusted based on the results.

The ACI mix design method is one of many methodsavailable but it is probably the most widely used sothat is the method we’ll use in this class. The methodinvolves ten steps outlined on the next page.

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Mix Design Steps1. Select the slump2. Select the NMAS3. Estimate the water and air contents4. Adjust the water content for aggregate shape5. Determine the required strength6. Select the w/cm ratio7. Calculate the cement weight8. Estimate the coarse aggregate content9. Calculate the fine aggregate content

10. Adjust for aggregate moisture and absorption

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ACI Mix Design

We’ll work through the mix design steps listed in the previous slide using an example for a typical concrete mix for a non-air-entrained concrete.

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2.90

Coarse aggregate = subangular crushed stone

Mix Design ExampleCoarse aggregate = subangular crushed stone

Nominal maximum aggregate size = 3/4"Design strength = 4500 psi

Specified slump = 1-2"

Coarse FineAggregate Aggregate

Unit weight (lb/ft3) = 101 106Bulk specific gravity (dry) = 2.574 2.548Bulk specific gravity (SSD) = 2.623 2.592Apparent specific gravity = 2.705 2.664Absorption capacity (%) = 1.9 1.7

Fineness modulus = 2.51 2.90CIVL 3137 5

Step 1: Select the slump

The choice of slump determines the workability ofthe mix. Workability encompasses a combination ofPCC properties that are related to the rheology of theconcrete mix: ease of mixing, ease of placing, easeof compacting, ease of finishing. You should aim forthe stiffest mix that will provide adequate placement.

The following table shows some typical slump rangesfor several different applications.

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Source: Design and Control of Concrete Mixtures (PCA, 2003)


For our mix design example, the slump has alreadybeen specified as 1-2".

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Step 2: Select the NMAS

The maximum aggregate size will affect parameterssuch as cement paste content, workability and strength.In general, the maximum aggregate size is limited bythe dimensions of the finished product and the roomavailable inside the formwork, taking into accountthings such as rebar. If the coarse aggregate is toolarge the concrete may be difficult to consolidate andcompact in the forms, resulting in a honeycombedstructure or large air pockets.

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narrowest dimensionNMAS 5

depth of slabNMAS 3

NMAS 0.75 clear space


For our mix design example, the nominal maximumaggregate size has already been specified as 3/4".

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Step 3: Estimate the water and air

The amount of mixing water basically determines theamount of cement paste in the mix. It depends on thedesired slump, the size and shape of the aggregateand the amount of air present in the mix. Some air(called entrapped air) is normal and is a consequenceof the mixing process. Admixtures can also be usedto introduce entrained air in order to enhance thefreeze/thaw durability of the concrete.

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The table on the next slide recommends the amountof water per cubic yard of concrete as a function ofthe desired slump and the NMAS. The top half of thetable is for non-air-entrained mixes and includes anestimate of the amount of entrapped air in the concrete.

The bottom half is for air-entrained mixes. It includestarget air contents based on the expected severity ofthe freeze/thaw exposure.

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For the ¾" NMAS in our mix design example, theamount of entrapped air is estimated as 2%. For thedesired slump of 1-2" the required water content isestimated to be 315 lb per cubic yard of cement.

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Questions to Ponder

Why does the amount of water required to obtain adesired slump decrease with increasing NMAS?

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Questions to Ponder

The amount of water largely determines the amountof cement paste in the mix. The amount of cementpaste needed to produce a workable concrete mixdepends in part on the surface area of the aggregateto be coated. As shown in the next slide, largeraggregate has a lower specific surface (surface areaper unit volume) so less cement paste is needed, thusless water is needed.

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Effect of NMAS on Paste Volume

surface area = 11 ft2 surface area = 22 ft2

10"

Questions to Ponder

A mix with a large NMAS may only require 30% byvolume of cement paste while a mix with a smallerNMAS may require 40% by volume of cement paste.The mix with the larger NMAS therefore requires25% less cement paste and thus 25% less water. Thisis illustrated in the next slide.

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Effect of NMAS on Paste Volume

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30%Cement

Paste

70%Aggregate

LargerNMAS

40%Cement

Paste

60%Aggregate Smaller

NMAS

Questions to Ponder

Why does the amount of entrapped air in a concretemix decrease with increasing NMAS?

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Questions to Ponder

The answer to this question is related to the previousquestion. The only place in the mix where there isentrapped air is in the cement paste. The air contentin the table is the amount of air per unit volume ofconcrete. If all of the entrapped air is in the cementpaste and there is less cement paste, it stands toreason that the air content of the concrete will belower.

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Questions to Ponder

Why does the target air content in an air-entrainedmix decrease with increasing NMAS?

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Questions to Ponder

The answer to this question is related to the previoustwo questions as well. The goal of air entrainment isto achieve a certain air content in the cement paste.If, for durability reasons, the required air content ofthe paste is the same in two mixes, but one requires25% more paste (due to a smaller NMAS), then thetarget air content of the concrete will automaticallybe higher as shown in the next slide.

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Air Content

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Concrete Air Content0.4 16% = 6.4%

Paste Air ContentAssume 16%

40%Cement

Paste

60%Aggregate Smaller

NMAS

Air Content

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30%Cement

Paste

70%Aggregate

Concrete Air Content0.3 16% = 4.8%

Paste Air ContentAssume 16%

LargerNMAS

Questions to Ponder

Why do you need less water in an air-entrained mixthan a non-air-entrained mix with the same NMAS?

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Questions to Ponder

The short answer is that cement paste with a higherair content takes up more space. Mix proportioningis about having the right volume proportions of thevarious ingredients, so less cement and water areneeded to produce the exact same volume of cementpaste. In our example, 280 lb of water will producethe same volume of air-entrained cement paste as isproduced by 315 lb of water in the non-air-entrainedcement paste.

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Step 4: Adjust for Aggregate Shape

An often overlooked part of the table used to estimatethe water content is the passage at the bottom, whichstates that the estimates are based on an assumptionof reasonably well-shaped angular coarse aggregate.

If you are using a rounded aggregate such as gravelrather than an angular aggregate such as crushed stoneyou need less water than is shown in the table. Thetable in the next slide estimates the adjustmentsneeded.

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Aggregate Shape

Water Reduction(pounds per cubic yard)

Crushed stone (angular) 0

Crushed stone (subangular) 20

Gravel (some crushed) 35

Gravel (well rounded) 45


The mix design example says the coarse aggregate is“subangular” so it is suggested that we reduce theamount of water by 20 lb/yd3, so instead of 315 lb/yd3

of water, we should start with

Ww = 315 – 20 = 295 lb/yd3

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Why does the water required to obtain a given slumpchange as a function of aggregate shape?

Questions to Ponder

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Aggregate Shape

Water Reduction(pounds per cubic yard)

Crushed stone (angular) 0

Crushed stone (subangular) 20

Gravel (some crushed) 35

Gravel (well rounded) 45

Questions to Ponder

Remember that the water content determines the pastecontent. Rounded aggregate has less surface area perunit volume of aggregate, as shown in the next slide,so you need less paste to coat the aggregate and thusless water.

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Minimizing Surface Area

surface area = 6.0 ft2/ft3 surface area = 4.8 ft2/ft3

Step 5: Determine Required Strength

As we said in the last lecture, the required strength ofthe concrete mix is not the same as the design strength.The design strength is the minimum strength that isrequired from a structural standpoint. Since concretestrength can vary greatly from one batch to the next,you need to build in a factor of safety to ensure thatmost, if not all, of the concrete exceeds the designstrength. If you don’t yet know the variability, thetable on the next slide estimates the overdesign youneed to build into the mix.CIVL 3137 38


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Required Average Compressive Strength When DataAre Not Available to Establish a Standard Deviation

Adapted from ASTM C94


Since the design strength in our mix design exampleis 4500 psi and we don’t yet know the variability ofour mix from one batch to the next, we need to add1200 psi to achieve an adequate factor of safety, sowe need to design our mix for a strength of

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cr cf f 1200 4500 1200 5700 psi

Step 6: Select the w/cm ratio

The water-cement ratio is correlated with strengthand durability. In general, lower water-cement ratiosproduce stronger, more durable concrete. If naturalpozzolans (such as fly ash) are used then the ratiobecomes a water-cementitious material ratio.

The following table relates the required 28-daycompressive strength (including the overdesign factor)to the water-cement ratio for both non-air-entrainedand air-entrained concrete mixes.

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crf


Since our required concrete strength is 5700 psi, wewill have to interpolate in the table to get the correctw/cm ratio. Our required strength is 70% of the wayfrom the 5000-psi entry to the 6000-psi entry so:

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w cm 0.48 0.7 0.41 0.48 0.43

Questions to Ponder

Why is the w/cm ratio different for air-entrainedconcrete compared to non-air-entrained concrete?

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Questions to Ponder

In a previous lecture, we said that entraining air toincrease freeze/thaw durability comes as a price. Asthe air content of the cement paste increases, theconcrete strength drops precipitously as shown in thenext slide. To compensate for the loss of strength, youneed to use a much lower w/cm ratio.

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Effect of Air Content on Strength

Step 7: Calculate the cement content

Now that we know the amount of water in the mixand the required w/cm ratio, we can calculate theamount of cement we need in the mix:

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WW = w/cm ratio

watercement

Step 7: Calculate the cement weight

Based on 295 lb of water and a 0.43 w/cm ratio, theamount of cement our design mix requires is

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295W = 6860.43

cement lb

Step 8: Estimate coarse aggregate

Selection of coarse aggregate content is empiricallybased on mixture workability. The following tableestimates the volume percentage of coarse aggregate(based on bulk volume) needed to produce concretewith a proper degree of workability for reinforcedconcrete construction. For things like pavement slabsthat don’t require as much workability, ACI allowsthe values to be increased by up to 10 percent.

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o

bb


The values in the table are called the b/bo factor. In anutshell, it tells you how big a box you would needto build to exactly contain all of the coarse aggregatein your mix (including all of the void spaces betweenthe aggregate particles).

As shown in the next slide, if you are trying to makea volume of concrete with dimensions 1×1×bo you’dneed to build a box with dimensions 1×1×b to holdall the coarse aggregate.

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What does b/bo represent?

Ratio of bulk aggregate volume (b)to bulk concrete volume (bo)

b

bo

1


The b/bo factors are a function of the NMAS of thecoarse aggregate and the fineness modulus of thefine aggregate. As we’ve said before, the larger theaggregate, the less cement paste is needed to coat thesurface area, so the more room there is for coarseaggregate. Also, as the fineness modulus of the sandincreases it becomes coarser and the blend of coarseand fine aggregate becomes less gap-graded. As aresult you need slightly more sand and less gravel inthe mix.CIVL 3137 59


Once you know how large your virtual box needs tobe, you can calculate the weight of coarse aggregateneeded to fill that box by multiplying the volume ofthe box by the dry-rodded unit weight of the coarseaggregate.

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bulk bulkgravel o concrete gravelW b b V γ


In our example, the fineness modulus of the sand is2.90, which is halfway between 2.80 and 3.00, so weinterpolate a b/bo value of 0.61 for a ¾” NMAS andcalculate the required coarse aggregate content fromits dry-rodded unit weight as

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3

gravelftW 0.61 27 3 3

lb101yd ft

31663 lb yd

Step 9: Estimate fine aggregate

ACI provides two different methods to estimate theamount of fine aggregate needed. The first method,the estimated weight method, uses typical values forthe unit weight of concrete mixes to determine howmuch the concrete should weigh once it’s mixed.

Once we’ve estimated the weight of all the otheringredients, whatever is still missing must be that ofthe sand.

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total cement gravel sand waterW W W W W

sand total cement gravel waterW W W W W

Estimated Weight Method

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Step 9: Estimate fine aggregateEstimated Weight Method

NMAS

(in) Non‐Air‐Entrained Concrete Air‐Entrained Concrete

⅜ 142.0 137.5

½ 144.0 139.0

¾ 146.5 141.5

1 148.5 143.5

1½ 151.0 146.0

2 153.0 147.5

3 155.5 150.0

6 157.5 152.0

First Estimate of Concrete Unit Mass (lb/ft3)


Based on our ¾" NMAS, the density of the concreteshould be 146.5 lb/ft3, so our concrete should weigh

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concrete 3

lb146.5ft

3ft27 33 3956 lb yd

yd


Questions to Ponder

Why does the density rise with increasing NMAS?

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NMAS

(in) Non‐Air‐Entrained Concrete Air‐Entrained Concrete

⅜ 142.0 137.5

½ 144.0 139.0

¾ 146.5 141.5

1 148.5 143.5

1½ 151.0 146.0

2 153.0 147.5

3 155.5 150.0

6 157.5 152.0

First Estimate of Concrete Unit Mass (lb/ft3)

Questions to Ponder

As we’ve said repeatedly, the larger the NMAS theless cement paste is needed to coat the surface areaof the aggregate. Since cement paste is less densethan a typical aggregate, a mix with more cementpaste will be less dense than a mix with less cementpaste.

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Questions to Ponder

So what is a typical density for cement paste?

To answer that, we’ll start with the observation thatthe volume of the cement paste is equal to the sum ofthe volumes of the cement and water (if we ignoreany entrapped air).

The weights of the cement and water can be foundby dividing their volumes by their specific gravitiesand the unit weight of water.

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Effect of NMAS on Unit Weight

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paste

paste w

WRD

water

water w

WRD

cement

cement w

WRD

water cement water cement

paste water cement

W W W WRD RD RD

paste water cementV V V

Questions to Ponder

If we assume a typical w/cm of 0.5 then the weightof the water is 0.5 times the weight of the cementand the total weight of the cement paste is 1.5 timesthe weight of the cement. As shown on the nextslide, this leads to a typical specific gravity of 1.83for the cement paste.

Aggregate typically has a specific gravity of 2.5-2.7,so cement paste is 2/3 to 3/4 as dense as aggregate.

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Effect of NMAS on Unit Weight

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cement1.5 W cement

paste

0.5 WRD

cement1.0 W1.00

3.15

Assume w/c = 0.5

pasteRD 1.83

aggregateRD 2.65 typical


If our concrete has a “typical” unit weight of 3956 lbper cubic yard of concrete then, using the estimatedweight method, the amount of sand that is needed tocomplete the mix design is

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3sandW 3956 686 1663 295 1312 lb yd



The estimated weight method is very approximatebecause it’s based on “typical” unit weights. A moreprecise method is the absolute volume method, whichdetermines the volume occupied by each ingredientbased on its bulk specific gravity (this is what ismeant by the absolute volume) then subtracts thosefrom 27 ft3 (1 yd3) to get the required volume of thesand. Since the entrapped or entrained air occupiessome of that volume, it needs to be included, too.

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total cement gravel sand water airV V V V V V

sand total cement gravel water airV V V V V V

Absolute Volume Method


In this approach, we use the bulk specific gravitiesof the aggregate to determine their absolute volumesbecause all of the water in the mix is supposed to bein the cement paste and not in the pervious pores ofthe aggregate. We will later add some water to themix to ensure the aggregate is SSD and doesn’t try toabsorb water from the cement paste.

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sand total cement gravel water airV V V V V V

gravelbulkgravel

cement watersand total air

w

W1 W WV V Vγ 3.1 G5 1.00

sand sabul

ndk

wsandGW V γ



We originally estimated that our mix would contain2% entrapped air, which is 0.54 ft3/yd3 of concrete, so

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sand1 686 1663

2.574295V 27 0.54

62.4 3.15 1.00


3 3sandV 27 18.57 0.54 7.89 ft yd


Now that we know the absolute volume of the sand,we can determine its weight from its specific gravity:

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3sandW 7.89 62.4 = 122 5.548 4 lb yd

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Step 10: Adjust for Moisture Content

The final step in the mix design (whether we used theabsolute volume or estimated weight method) is to(1) add additional water to the mix to make sure theaggregate is saturated and doesn’t absorb water fromthe cement paste, and (2) adjust the weights of theaggregate and the mixing water to account for thefact that the aggregate stockpiles at the batch plantwill not be oven-dry.

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1. Increase Wwater by an amount equal to the weight of water needed to saturate the fine and coarse aggregate.

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Since we did our calculations based on bulk OD specific gravity …

… we‘ve assumed the pervious pores are filled with air.

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If we don’t add enough extra water to fill those pervious pores …

… the aggregate will suck water out of the cement paste.


The amount of water needed to saturate the aggregateis just the product of the aggregate weight and theaggregate absorption:

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3waterW 295 53 348 lb yd

3waterW 0.019 1663 0.017 1254 53 lb yd


So, based on the absolute volume method calculations,our “laboratory” mix design (i.e., what we’d make inthe laboratory using oven-dry aggregate) is

Wwater = 348 lb/yd3

Wcement = 686 lb/yd3

WOD gravel = 1663 lb/yd3

WOD sand = 1254 lb/yd3

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2. Increase Wsand and Wgravel to account for the current moisture contents of the aggregate in the batch plant stockpiles.

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If our mix design calls for 1000 lb of dry aggregate …

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… but the moisture content is currently 10% …

… we have to weigh up 1000 (1.10) = 1100 lb of moist aggregate.

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For our example, assume the sand stockpile has amoisture content of 6.2% and the gravel stockpilehas a moisture content of 2.1% on the day we aregoing to batch our concrete mix. Then

3wet sandW 1.062 1254 1332 lb yd

3wet gravelW 1.021 1663 1698 lb yd

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2. Increase Wsand and Wgravel by an amount equal to the moisture contents of the aggregate stockpiles.

3. Decrease Wwater by the same amount you increased Wsand and Wgravel.

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Since we’ve weighed up 1000 lb of dry aggregate + 100 lb of water …

… we have to reduce the amount of water we add from the faucet by 100 lb.


Since Mother Nature is providing some of the waterneeded in the mix, we can reduce the amount we addfrom the faucet by a like amount:

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3waterW 348 113 235 lb yd

3waterW 1698 1663 1332 1254 113 lb yd


So, our “field” mix (i.e., what we’d make in the fieldtoday using aggregate in its current moisture state) is

Wwater = 235 lb/yd3

Wcement = 686 lb/yd3

Wwet gravel = 1663 lb/yd3

Wwet sand = 1254 lb/yd3

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